[/quote]
,
and 
. Prove that
Where 
such that: 
.
be a cyclic quadrilateral, and let 
,
,
,
be the midpoints of 
,
,
,
respectively. Let 
,
,
and 
be the orthocenters of triangles 
,
,
and 
,
have the same area.
be an acute triangle with incentre 
and 
.
and 
intersect the circumcircle of 
and 
,
and 
such that 
and 
are parallelograms (with 
,
)
be the point of intersection of lines 
and 
.
,
![\[
(AB + CD)^2 + (AD + BC)^2 = (AC + BD)^2,
\]](http://latex.artofproblemsolving.com/7/d/a/7daf2d6ac00609b5252b2b8b05d96a5ca0952a94.png)
then 
touch sides 
at 
,
lie on the line through 
perpendicular to 
be the intersections of 
with 
,
be the projections of 
onto line 
be the second intersections of 
with the incircle 
and 
.
bisects segment 
.
,
-
,
.
at 
.
parallel to 
respectively, so that 
meets 
.
is constant.
.
to 
respectively.
and 
as the reflections of 
be the circumcircle of 
.
with 
the intersection of ray 
with 
is the circumcenter of 
can be inscribed within a circle.
![\[ a = b\cos C + c\cos B. \]](http://latex.artofproblemsolving.com/7/e/d/7ed699f18ede191e5b89bc79d7f0592294347818.png)
be the foots of the perpendiculars from 
,
and 
intersect at 
and 
intersect at 
are parallel.
.![\[
2\angle AED-\angle DCB =180^\circ.
\]](http://latex.artofproblemsolving.com/b/d/0/bd01635e002970e841cef9dcb98b154ad6a7a202.png)
![\[
\alpha = |\angle ADB - \angle ACB| = |\angle DAC - \angle DBC|
\quad\text{and}\quad
\beta = |\angle ABD - \angle ACD| = |\angle BAC - \angle BDC|.
\]](http://latex.artofproblemsolving.com/3/e/e/3eef0425824a79ba8615288dcfe2df79fca048b1.png)
Prove that![\[
AC \cdot BD = AD \cdot BC \cos\alpha + AB \cdot CD \cos\beta.
\]](http://latex.artofproblemsolving.com/d/4/a/d4adc5d510093da3520b26cbce172cf791b2847a.png)
be the sum of two opposite angles. Prove that![\[
AC^2 \cdot BD^2 = AB^2 \cdot CD^2 + AD^2 \cdot BC^2 - 2AB \cdot CD \cdot AD \cdot BC \cos\theta.
\]](http://latex.artofproblemsolving.com/4/2/4/424dfbb285f46ccc5524b439028bd2ef54569c03.png)
Y by Adventure10
It is known that: If 
are nonnegative real numbers such that 
then
Now, please try with my improvement (which is so much sharper): Let be positive real numbers such that 
Prove that
(This is sharper because it can be written as
![$ab^2+bc^2+ca^2 +abc \left[ 1+\frac{1}{6}(a^2+b^2+c^2-ab-bc-ca)\right] \le 4.$](//latex.artofproblemsolving.com/3/d/a/3da9073289c5daa60333de05bec6783bb2cde721.png)
)
Notice that in the above inequality, the constant
is the best possible.
are nonnegative real numbers such that 
then
Now, please try with my improvement (which is so much sharper): Let
be positive real numbers such that 
Prove that
(This is sharper because it can be written as
![$ab^2+bc^2+ca^2 +abc \left[ 1+\frac{1}{6}(a^2+b^2+c^2-ab-bc-ca)\right] \le 4.$](http://latex.artofproblemsolving.com/3/d/a/3da9073289c5daa60333de05bec6783bb2cde721.png)
)Notice that in the above inequality, the constant
is the best possible.
fro
,
it is ture when 
t
it is ture
then Both sides square
.
